Question 636348
what is the lowest-degree polynomial with integer coefficients and the roots 6, i, and -i?
<pre> 
It doesn't work to guess.

Set x = to each root:

     x=6,   x=i,   x=-i

Get 0 on the right side of each:

   x-6=0  x-i=0, x+i=0

Indicate the multiplication of all
three left sides and set equal to 0:

    (x-6)(x-i)(x+i) = 0

Multiply two of them (the easiest is the
second and third, since they are conjugates:

       (x-6)(x²-i²) = 0

Use the fact that i²=-1 to replace i²

     (x-6)[x²-(-1)] = 0

        (x-6)(x²+1) = 0

Multiply the left side out:

         x³+x-6x²-6 = 0

Arrange in descending powers of x:

         x³-6x²+x-6 = 0  

So you see that the answer is c., not d.

Edwin</pre>